At 8/27/2007 20:52, you wrote: >Yes your right VSWR is the ratio between Vmax and Vmin, node and anodes, >of the interference pattern caused by standing waves. Even still there is >a point where the voltage is at a minimum on the line. What happens if >that point is at the transmitters output... does it help keep the heat >down in the transmitter due to high SWR? > > > >It doesn t matter where the min and max are on the line. The same amount >of reflected power will be seen at any point. Reflected power does NOT get >back into the transmitter. It gets re-reflected back towards the antenna >when it reaches the transmitter circuits.
I don't buy into this. In order for reflected power to not be absorbed by the TX, it would have to appear totally reactive. Although I've never measured one, I don't believe that's the case. >If you have two watt meters and an antenna matching device you can put one >wattmeter between the transmitter and the matching device and tune it for >minimum reflected power on the first meter. Then with a second meter >between the tuner and the mismatched load you can see the second wattmeter >that is reading the reflected power. The second wattmeter will have a >higher forward power reading than the first due to the added re-reflected >power. This doesn't sound right either, as there should be no reflected power at the antenna if it's been matched further down the line. The tuner would be adjusted so as to create a conjugate impedance of the antenna at the end of the feeding coax, thus eliminating the mismatch. My guess is that the higher power reading on the wattmeter is due to the weird impedances it's seeing on both its input & output. Bob NO6B
